Giải thích các bước giải:
\(\begin{array}{l}
{\left( {a + b + c} \right)^2} = 3ab + 3bc + 3ca\\
\Leftrightarrow {\left[ {\left( {a + b} \right) + c} \right]^2} = 3ab + 3bc + 3ca\\
\Leftrightarrow {\left( {a + b} \right)^2} + 2.\left( {a + b} \right).c + {c^2} = 3ab + 3bc + 3ca\\
\Leftrightarrow {a^2} + 2ab + {b^2} + 2ac + 2bc + {c^2} = 3ab + 3bc + 3ca\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 3ab + 3bc + 3ca\\
\Leftrightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca - 3ab - 3bc - 3ca = 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0\\
\Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0\\
\Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) = 0\\
\Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\,\,\,\,\left( 1 \right)\\
{\left( {a - b} \right)^2} \ge 0,\,\,\,\forall a,b\\
{\left( {b - c} \right)^2} \ge 0,\,\,\,\forall b,c\\
{\left( {c - a} \right)^2} \ge 0,\,\,\,\forall c,a\\
\Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \ge 0,\,\,\,\forall a,b,c\\
\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}
{\left( {a - b} \right)^2} = 0\\
{\left( {b - c} \right)^2} = 0\\
{\left( {c - a} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a - b = 0\\
b - c = 0\\
c - a = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = b\\
b = c\\
c = a
\end{array} \right. \Leftrightarrow a = b = c
\end{array}\)
